Why does the period of a simple pendulum increase as the oscillation angle $\theta$ increases?

Question

I understand that there are differential equations that model the pendulum. I also know that, the period of the pendulum is given by $2\pi/\omega$. However, in the case of the simple pendulum, angular velocity, $\omega$ is changing. How should I approach this problem if I want to find the period?

Answer 1

Start with conservation of energy $$ E=\frac{m}{2}\ell^2\dot{\theta}^2+mg\ell(1-\cos\theta)\, . \tag{1} $$ The potential energy $mg\ell(1-\cos\theta)$ is not exactly quadratic in $\theta^2$, with the result that the period will generally depend on the amplitude of the motion. This is a pretty generic feature of all potentials, and it’s only for the quadratic potential and other specialized situation (see the article on the cycloid pendulum ) that the period is independent of the amplitude (the fancy word is isochronous).

To make progress, the trick is to evaluate $E$ at a turning point, i.e. this value $\theta_0$ where the pendulum has maximum amplitude, zero velocity, and reverses course. Since $\dot{\theta}=0$ there, we easily obtain \begin{align} mg\ell-mg\ell\cos\theta_0&=\frac{m}{2}\ell^2\dot{\theta}^2+mg\ell-mg\ell\cos\theta\, ,\\ \frac{g}{\ell}\left(\cos\theta-\cos\theta_0\right)&=\frac{1}{2}\dot{\theta}^2\, . \tag{2} \end{align} If you think of $\dot{\theta}$ as $\omega$, then this shows that $\omega$ is not constant along the trajectory, but this is expected since at the turning point the angular velocity $\omega=0$.

The next step is to reorganize (2) as \begin{align} dt=\frac{d\theta}{(2g/\ell)^{1/2}\sqrt{\cos\theta-\cos\theta_0}}\, . \end{align}

To continue, we need notions of integral calculus and integrate over one-quarter period, which means the pendulum goes from the equilibrium $\theta=0$ to the turning point at $\theta_0$: \begin{align} \frac{T}{4}=\sqrt{\frac{\ell}{2g}}\int_0^{\theta_0} \frac{d\theta}{\sqrt{\cos\theta-\cos\theta_0}}\, . \end{align} This last integral does not have a closed form in terms of simple functions. Nevertheless, we can expand the denominator in powers of $\theta$ and $\theta_0$.

This yields \begin{align} \cos\theta-\cos\theta_0&=\frac{1}{2}\, \left({\theta_0}^2-\theta^2\right)-\frac{1}{24}\, \left({\theta_0^4}-\theta^4\right)+\ldots \, ,\\ \left(\cos\theta-\cos\theta_0\right)^{-1/2}&\approx \frac{\sqrt{2}}{\sqrt{\theta^2_0-\theta^2}} +\frac{\theta_0^2+\theta^2}{12\sqrt{2}\sqrt{\theta^2_0-\theta^2}}\, , \tag{3}\\ \frac{T}{4}&\approx\sqrt{\frac{\ell}{g}}\displaystyle\int_0^{\theta_0} \frac{d\theta}{\sqrt{\theta^2_0-\theta^2}} \left(1+\frac{\theta_0^2}{24}+\frac{\theta^2}{24}\right)\, . \end{align} Once you’ve past this painful stage (I’ve only carried terms to ${\cal O}(\theta^4)$), you can now integrate term by term using the trig substitution $\theta=\theta_0\sin\xi$ eventually reaching \begin{align} T&=2\pi\,\sqrt{\frac{\ell}{g}}\, \left(1+\frac{\theta_0^2}{16}\right)\, . \end{align} The factor $2\pi \sqrt{\ell/g}$ is the usual period of the pendulum. The correction term is thus $\theta_0^2/16$. This correction term depends on the amplitude of the motion.

For small angles, say $\theta_0=\pi/12$ (or $15^\circ$), this correction is thus approximately $4\times 10^{-3}$ of the purely harmonic $2\pi \sqrt{\ell/g}$ period.

You get the approximate but well-known result $2\pi\sqrt{\ell/}$, independent of the amplitude $\theta_0$, by keeping only the first term on the right hand side of (3) and proceeding with the integration. This is the “small angle’’ limit where $\cos\theta-\cos\theta_0\approx \frac{\theta_0^2}{2}-\frac{\theta^2}{2}$.

Answer 2

The differential equation of the simple pendulum gets a little messy, since it cannot be solved in closed form using elementary functions. The standard analytical solution involves a complete elliptic integral of the first kind.

However, as that Wikipedia article mentions, there's a lovely iterative algorithm originally due to Gauss which uses the arithmetic-geometric mean so it converges rapidly, and is ideal for computer calculations. See Carlson symmetric form for more info & links to modern articles by Bille C. Carlson on this topic.

Several years ago, "Time hacker" Tom van Baak wrote an excellent series of articles about precision pendulums. In particular, see A New and Wonderful Pendulum Period Equation, which gives a simple explanation of the AGM formula.

From Tom's PDF, the equation of the period $T$ of a simple pendulum of length $L$ can be written as

$$T = 2\pi \sqrt\frac{L}{g}\left(1 + CE \right)$$

where $g$ is the local acceleration due to gravity, and $CE$ is a small circular error correction term. Its value for a given angle $\theta$ (measured from the vertical, as usual) can be found using a series in $\theta^2$, as shown in ZeroTheHero's answer.

If we use the AGM, that equation becomes

$$T = 2\pi \sqrt\frac{L}{g}\left(1/\mathrm{AGM}\left(1, \cos\left(\theta/2\right)\right)\right)$$

The Wikipedia article on the AGM gives its definition, which I repeat here, for convenience.

$$\begin{align} a_{n+1} & = (a_n + g_n) / 2\\ g_{n+1} & = \sqrt{a_n g_n} \end{align}$$

Here's a comparison of the corrections using the AGM and the $\theta^2$ series.

The 1st row of each block shows the angle in degrees and the AGM correction, i.e., $1/\mathrm{AGM}(1, \cos(\theta/2))-1$. The 2nd row has a $\theta^2$ & a $\theta^4$ term. The 3rd row just has the $\theta^2$ term. The coefficient for the $\theta^4$ term (taken from Tom's article) is 11 / 3072.

 0: 0.000000000000000
    0.000000000000000
    0.000000000000000

 1: 0.000019038921006
    0.000019038920999
    0.000019038588737

 2: 0.000076159671567
    0.000076159671142
    0.000076154354947

 3: 0.000171374216705
    0.000171374211869
    0.000171347298630

 4: 0.000304702506088
    0.000304702478911
    0.000304617419787

 5: 0.000476172485987
    0.000476172382295
    0.000475964718417

 6: 0.000685820116037
    0.000685819806339
    0.000685389194520

 7: 0.000933689390824
    0.000933688609653
    0.000932890848097

 8: 0.001219832366326
    0.001219830625143
    0.001218469679147

 9: 0.001544309191237
    0.001544305660004
    0.001542125687670

10: 0.001907188143217
    0.001907181495727
    0.001903858873667

11: 0.002308545670118
    0.002308533888095
    0.002303669237137

12: 0.002748466436233
    0.002748446567184
    0.002741556778080

13: 0.003227043373625
    0.003227011237362
    0.003217521496497

14: 0.003744377738611
    0.003744327577293
    0.003731563392387

15: 0.004300579173465
    0.004300503239929
    0.004283682465751

16: 0.004895765773430
    0.004895653852519
    0.004873878716587

17: 0.005530064159106
    0.005529903016604
    0.005502152144897

18: 0.006203609554339
    0.006203382308017
    0.006168502750681

19: 0.006916545869678
    0.006916231276885
    0.006872930533938

20: 0.007669025791545
    0.007668597447627
    0.007615435494668

I generated those figures using Sage / Python. Sage has a built-in AGM function. You can play with the script here.

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The AGM formulae for various elliptic integrals deserve to be more well-known, IMHO. They're relatively obscure because the AGM has no closed form in elementary functions, and it was tedious to calculate before the computer age, since it involves both addition & multiplication. In the old days, logarithms were generally used for multiplication, so iterating the AGM would require constantly going back & forth between logarithms and antilogarithms.

Answer 3

I think the other answers are answering a more difficult question: this is about a simple pendulum so there is no need to go nonlinear.

The questioner has been fooled by the ambiguity in our use of $\omega$. It is the 'angular' velocity of the wave, and for a mass on a spring we happily write $x=A \sin(\omega t)$ as equivalent to $x=A \sin(2 \pi t /T)$ and it is obvious that the 'angle' $\omega t$ which is an argument to the sine function is not one which can be measured with a protractor.

It gets confusing when we switch to the pendulum, which is mathematically the same except that the variable is called $\theta$ rather than $x$ so we write $\theta = A \sin(\omega t)$. From which, trivially, ${ d \theta \over dt}=\omega A cos (\omega t)$. But the ${d \theta \over dt}$ on the left is not the same as the $\omega$ on the right.

There are two different angular velocities in the simple pendulum. One is constant and equal to $2\pi/T$, the other is variable and linked to a real protractor-measurable angle. You can call either =of these $\omega$ if you choose. But not both.